Comments

Author: sherm1

multibody/tree/mobilized_body.h

@@ -176,7 +176,7 @@ where Hv_FM = ∂p_FM/∂q ⋅ N is the last three rows in H_FM.
 
 The angular velocity w_FM component of the spatial velocity V_FM can be
 related to the time derivative of the rotation matrix R_FM in X_FM. The
-time derivative of a rotation matrix R is Ṙ = [ω]R where [ω] is the 
+time derivative of a rotation matrix R is Ṙ = [ω]R where [ω] (=ṘRᵀ) is the
 skew-symmetric cross product matrix formed from the angular velocity ω.
 We have the identity Rᵀ[ω]R=[Rᵀω] for orthogonal R. Premultiplying the Ṙ
 equation by Rᵀ then gives [Rᵀω] = RᵀṘ which we can use to write
@@ -194,6 +194,17 @@ or          [Hw_FM ⋅ v] = ∂R_FM/∂q ⋅ N ⋅ v ⋅ R_FMᵀ
 
 TODO(alejandro) Can the above be simplified to get rid of v to provide a
  more-direct relationship between Hw_FM and ∂R_FM/∂q as we have for translation?
+ Here is the best I've come up with so far:
+
+Dropping the frames for clarity, we can write the RHS as
+ | ∂R₀Rᵀ ∂R₁Rᵀ ∂R₂Rᵀ ∂R₃Rᵀ | ⋅ N ⋅ v
+where ∂Rᵢ≜∂R/∂qᵢ.
+
+That matrix can be viewed as a 1 x 4 hypermatrix (each entry is 3x3)
+compatible with the 4 x 3 N matrix, yielding a 1 x 3. The final multiply by
+v (3 x 1) yields a single (1x1) 3x3 matrix, matching the LHS. That 1x3
+matrix | ∑Nᵢ₀∂RᵢRᵀ ∑Nᵢ₁∂RᵢRᵀ ∑Nᵢ₂∂RᵢRᵀ | consists of the three columns of
+Hw_FM (each a 3 vector) written as a skew-symmetric matrix.
 
 
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